# Strong law of large numbers and divergence of partial sum

Let $$(\Omega, \mathcal{F}, \mathbb{P})$$ be a probability space. Let $$(X_n)_{n\in \mathbb{N}}$$ be a sequence of real random variables in $$L^1(\mathbb{P})$$. Each $$X_n$$ has a mean $$c>0$$.

Suppose that $$(X_n)$$ satisfies the strong law of large numbers, i.e., (I omit a.s. when it's clear) $$\limsup_n |n^{-1} S_n- c|=0,$$ where $$S_n = \sum_{i=1}^n X_i$$ .

Is it true that $$\liminf_n S_n =\infty$$? This is intuitively true because $$n^{-1} S_n$$ concentrates around $$c>0$$ for large $$n$$, but I'm not sure how to show this.

I can show that $$\limsup_n S_n = \infty$$. If $$\limsup S_n=k <\infty$$, then $$\frac{S_{n}}{n}\le\frac{\limsup S_{n}}{n}=\frac{k}{n}\to 0,$$ which is a contradiction.

• The assertion $S_n \leq \limsup S_n$ is not correct. Aug 15, 2023 at 22:08
• Ah I see. Then I don't know where to start for bot limsup and liminf... Aug 15, 2023 at 22:16

Hint: The answer is YES. Almost surely $$\frac {S_n} n >\frac c 2$$ for $$n$$ sufficiently large. This implies $$S_n >\frac {nc} 2$$ for $$n$$ sufficiently large, so $$S_n \to \infty$$ with probability $$1$$.
• I understand the implication but I'm not sure how to prove $n^{-1}S_n > c/2$ for $n$ sufficiently large. For example, if $n^{-1}S_n=(-1)^{n}$ and $c=1$, then the assertion is not true. For this particular sequence, it would be impossible though, as $X_n$ would not be in $L^1$. However, in general, how can I use the assumption to show the assertion? I understand that if a.s. convergence is defined with lim instead of limsup, it is trivial. Aug 16, 2023 at 21:39
• If $a_n \geq 0$ and $\lim \sup a_n=0$ then $a_n \to 0$. So $\frac 1 n S_n \to 0$ almost surely. @keepfrog Aug 16, 2023 at 23:13
• In your example, the hypothesis is not satisfied. Perhaps, you are confusing $\lim \sup$ with $\lim \inf$ @keepfrog Aug 16, 2023 at 23:20